What is the voltage in a circuit that has a current of 10.0 amps and a resistance of 28.5 ohms?

The diagram below shows a person using a device called a jetpack. Water is forced downwards from the jetpack and produces an upw

klasskru [66]

Answer:

The condition necessary for the person to be able to remain stationary in mid-air is that there must be an upward force also referred to as THRUST

Explanation:

Thrust in Physics is defined as the force that propels, forces or pushes the mass of a certain object in a specified or particular direction.

Thrust can defined in the terms of a Jet pack can be defined as the force that is required to propel the mass of a person in an upward direction.

Based on the diagram that we have been shown the question, we can see that water is the fuel used to proper the user or person upwards hence, the jet pack used in the question is an HydroJet pack

The condition necessary for the person to be able to remain stationary in mid-air is that the propelling force called the thrust must be upwards such that it can overcome these 4 things:

a) Gravitational force

b) Mass of the jet pack user

c) Mass of the jet pack itself

d) The water which serves as fuel for the jet pack.

5 0

2 years ago

A closed, rigid container holding 0.2 moles of a monatomic ideal gas is placed over a Bunsen burner and heated slowly, starting

Georgia [21]

Answer:

a) 2250 J

b) 0 J

c) 2250 J

Explanation:

a) Since, the process is isochoric

the change in internal energy

$\Delta U = n C_v(T_f-T_i)$

Here, n = 0.2 moles

Cv = 12.5 J/mole.K

We have to find T_f so we can use gas equation as

$\frac{P_1V_1}{P_2V_2} =\frac{T_i}{T_f}\\Since, V_1=V_2 [isochoric/process]\\\Rightarrow \frac{P_{atm}}{4P_{atm}} = \frac{300}{T_f} \\\Rightarrow T_f = 1200 K$

So, $\Delta U= 0.2\times12.5(1200-300)\\=2250 J$

b) Since, the process is isochoric no work shall be done.

c) By first law of thermodynamics we have

$\Delta U = Q-W\\Since, W = 0\\\Delta U = Q\\Therefore, Q = 2250 J$

Since, Q is positive 2250 J of heat will flow into the system.

6 0

2 years ago

In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is

Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

$\theta$ = Angle = 20°

y = -20 cm

Velocity components

$u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s$

$u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s$

Acceleration components

$a_x=0$

$a_y=-9.81\ m/s^2$

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0$

$t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629$

Time taken is 0.26088 seconds

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m$

The distance the beetle travels on the ground is 0.3677181864 m

6 0

2 years ago

Starting at point 0, you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees. What is your heigh

ira [324]

Answer:

43.58 m

Explanation:

If you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees

Using trigonometry ratio

Sin 5 = opposite/hypothenus

Where the hypothenus = 500m

Opposite = height h

Sin 5 = h/500

Cross multiply

500 × sin 5 = h

h = 500 × 0.08715

h = 43.58m

Therefore, the height above the starting point is equal to 43.58m

5 0

2 years ago

A simple pendulum 0.64m long has a period of 1.2seconds. Calculate the period of a similar pendulum 0.36m long in the same locat

weqwewe [10]

The period of the second pendulum is 0.9 s

Explanation:

The period of a simple pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity at the location of the pendulum

For the first pendulum, we have

L = 0.64 m

T = 1.2 s

Therefore we can find the value of g at that location:

$g=(\frac{2\pi}{T})^2 L=(\frac{2\pi}{1.2})^2 (0.64)=17.5 m/s^2$

Now we can find the period of the second pendulum at the same location, which is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where we have

L = 0.36 m (length of the second pendulum)

$g=17.5 m/s^2$

Substituting,

$T=2\pi \sqrt{\frac{0.36}{17.5}}=0.9 s$

#LearnwithBrainly

8 0

2 years ago